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Introduction A Closer Look at Graphing Translations Definition of a Circle Definition of Radius Definition of a Unit Circle Deriving the Equation of a Circle A Closer Look at Graphing Translations Definition of a Circle Definition of Radius Definition of a Unit Circle Deriving the Equation of a Circle The Distance Formula Moving The Center Completing the Square It is a Circle if? Conic Movie Conic Collage EndNext Geometers Sketchpad: Cosmos Geometers Sketchpad: Headlights Projectile Animation Planet Animation Plane Intersecting a Cone Animation Footnotes Skip Intro

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Next ParabolaCircleEllipseHyperbola Quadratic Relations PreviousMain MenuEnd

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The quadratic relations that we studied in the beginning y = Ax 2 + Dx + F of the year were in the form of y = Ax 2 + Dx + F, where A, D, and F stand for constants, and A ≠ 0. This quadratic relation is a parabola, and it is the only one that can be a function. It does not have to be a function, though. A parabola is determined by a plane intersecting a cone and is therefore considered a conic section. NextPreviousMain MenuEnd

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The general equation for all conic sections is: Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 where A, B, C, D, E and F represent constants y 2 xy When an equation has a “y 2 ” term and/or an “xy” term it is a quadratic relation instead of a quadratic function. and where the equal sign could be replaced by an inequality sign. NextPreviousMain MenuEnd

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A Closer Look at Graphing Conics Plot the graph of each relation. Select values of x and calculate the corresponding values of y until there are enough points to draw a smooth curve. Approximate all radicals to the nearest tenth. 1)x 2 + y 2 = 25 2)x 2 + y 2 + 6x = 16 3)x 2 + y 2 - 4y = 21 4)x 2 + y 2 + 6x - 4y = 12 5)Conclusions NextPreviousMain MenuLast ViewedEnd

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x 2 + y 2 = 25 x = y 2 = y 2 = 25 y 2 = 16 2 There are 2 points to graph: (3,4) (3,-4) y = ± 4 NextPreviousMain MenuA Close Look at GraphingEnd

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Continue to solve in this manner, generating a table of values ±3 -3±4 -2±4.6 ±4.9 0±5 1±4.9 2±4.6 3±4 4±3 50 NextPreviousMain MenuA Close Look at GraphingEnd

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x 2 + y 2 = 25 Graphing x 2 + y 2 = 25 x y NextPreviousMain MenuA Close Look at GraphingEnd

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x 2 + y 2 + 6x = 16 x = y 2 +6(1) = y 2 +6 = 16 y 2 = 9 2 There are 2 points to graph: (1,3) (1,-3) y = ± 3 NextPreviousMain MenuA Close Look at GraphingEnd

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Continue to solve in this manner, generating a table of values ±3 -6±4 -5±4.6 -4±4.9 -3±5 -2±4.9 ±4.6 0±4 1± ±3 -6±4 -5±4.6 -4±4.9 -3±5 -2±4.9 ±4.6 0±4 1±3 20 NextPreviousMain MenuA Close Look at GraphingEnd

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-80 -7±3 -6±4 -5±4.6 -4±4.9 -3±5 -2±4.9 ±4.6 0±4 1±3 20 x 2 + y 2 + 6x = 16 Graphing x 2 + y 2 + 6x = 16 NextPreviousMain MenuA Close Look at GraphingEnd

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x 2 + y 2 - 4y = 21 x = y 2 - 4y = y 2 - 4y = 21 y 2 - 4y -12 = 0 2 There are 2 points to graph: (3,6) (3,-2) y = 6 and y = -2 (y - 6)(y + 2) = 0 y - 6 = 0 and y + 2 = 0 NextPreviousMain MenuA Close Look at GraphingEnd

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Continue to solve in this manner, generating a table of values. 0-3, 7 3-2, 6 4-1, , , , , , , , 7 3-2, 6 4-1, , , , , , , 6.9 NextPreviousMain MenuA Close Look at GraphingEnd

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x 2 + y 2 - 4y = 21 Graphing x 2 + y 2 - 4y = , 7 3-2, 6 4-1, , , , , , , 6.9 NextPreviousMain MenuA Close Look at GraphingEnd

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x 2 + y 2 + 6x - 4y = 12 x = y 2 + 6(1) - 4y = y y = 12 y 2 - 4y - 5 = 0 2 There are 2 points to graph: (1,5) (1,-1) y = 5 and y = -1 (y - 5)(y + 1) = 0 y - 5 = 0 and y + 1 = 0 NextPreviousMain MenuA Close Look at GraphingEnd

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Continue to solve in this manner, generating a table of values , , , , , , , , , , , , , , , , , , NextPreviousMain MenuA Close Look at GraphingEnd

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x 2 + y 2 + 6x - 4y = 21 Graphing x 2 + y 2 + 6x - 4y = , , , , , , , , , NextPreviousMain MenuA Close Look at GraphingEnd

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x 2 + y 2 = 25 x 2 + y 2 + 4x = 16 x 2 + y 2 - 6y = 21 x 2 + y 2 + 4x - 6y = 12 What conclusions can you draw about the shape and location of the graphs? NextPreviousMain MenuA Close Look at GraphingEnd

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What conclusions can you draw about the shape and location of the graphs? x 2 + y 2 = 25 All of the graphs are circles. x 2 + y 2 = 25 x 2 + y 2 + 4x = 16 As you add an x-term, the graph moves left. x 2 + y 2 + 4x = 16 x 2 + y 2 - 6y = 21 As you subtract a y-term, the graph moves up. x 2 + y 2 - 6y = 21 x 2 + y 2 + 4x - 6y = 12 You can move both left and up. x 2 + y 2 + 4x - 6y = 12 NextPreviousMain MenuA Close Look at GraphingEnd

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circle center A circle is the set of all points in a plane equidistant from a fixed point called the center. Center Circle Center GEOMETRICAL DEFINITION OF A CIRCLE NextPreviousMain MenuEnd

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radius A radius is the segment whose endpoints are the center of the circle, and any point on the circle.radius radius NextPreviousMain MenuEnd

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A unit circle is a circle with a radius of 1 whose center is at the origin. It is the circle on which all other circles are based. A unit circle is a circle with a radius of 1 whose center is at the origin. r = 1 A unit circle is a circle with a radius of 1 whose center is at the origin. Center: (0, 0) unit circleradius of 1center is at the origin A unit circle is a circle with a radius of 1 whose center is at the origin. NextPreviousMain MenuEnd

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radius The equation of a circle is derived from its radius. NextPreviousMain MenuEndLast Viewed

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distance formula the equation for the circle Use the distance formula to find an equation for x and y. This equation is also the equation for the circle. NextPreviousMain MenuDeriving the Equation of a CircleEnd

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THE DISTANCE FORMULA (x 2, y 2 ) (x 1, y 1 ) NextPreviousMain MenuDeriving the Equation of a CircleEnd

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Deriving the Equation of a Circle (x, y) (0, 0) rradius Let r for radius length replace D for distance. r 2 = x 2 + y 2 Is the equation for a circle with its center at the origin and a radius of length r. NextPreviousMain MenuEnd

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The unit circle therefore has the equation: x 2 + y 2 = 1 (x, y) (0, 0) r = 1 NextPreviousMain MenuEndDeriving the Equation of a Circle

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x 2 + y 2 = 4 r = 2 If r = 2, then (x, y) (0, 0) r = 2 NextPreviousMain MenuDeriving the Equation of a CircleEnd

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In order for a satellite to remain in a circular orbit above the Earth, the satellite must be 35,000 km above the Earth. Write an equation for the orbit of the satellite. Use the center of the Earth as the origin and 6400 km for the radius of the earth. (x - 0) 2 + (y - 0) 2 = ( ) 2 x 2 + y 2 = 1,713,960,000 NextPreviousMain MenuDeriving the Equation of a CircleEnd

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What will happen to the equation if the center is not at the origin? (1,2) r = 3 (x,y) NextPreviousMain MenuEndLast Viewed

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No matter where the circle is located, or where the center is, the equation is the same. (h,k)(x,y)r (h,k) (x,y) r (h,k)(x,y)r (h,k) (x,y) r (h,k)(x,y)r NextPreviousMain MenuMoving the CenterEnd

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(x,y) (x, y) Assume (x, y) are the coordinates of a point on a circle. (h,k) (h, k) The center of the circle is (h, k),r and the radius is r. (x - h) 2 + (y - k) 2 = r 2 Then the equation of a circle is: (x - h) 2 + (y - k) 2 = r 2. NextPreviousMain MenuMoving the CenterEnd

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(x - h) 2 + (y - k) 2 = r 2 (x - 0) 2 + (y - 3) 2 = 7 2 (x) 2 + (y - 3) 2 = 49 center at (0, 3) radius of 7 Write the equation of a circle with a center at (0, 3) and a radius of 7. NextPreviousMain MenuMoving the CenterEnd

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Find the equation whose diameter has endpoints of (-5, 2) and (3, 6). First find the midpoint of the diameter using the midpoint formula. MIDPOINT This will be the center. NextPreviousMain MenuMoving the CenterEnd

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Find the equation whose diameter has endpoints of (-5, 2) and (3, 6). Then find the length distance between the midpoint and one of the endpoints. DISTANCE FORMULA This will be the radius. NextPreviousMain MenuMoving the CenterEnd

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Find the equation whose diameter has endpoints of (-5, 2) and (3, 6). Therefore the center is (-1, 4) The radius is (x - -1) 2 + (y - 4) 2 = (x + 1) 2 + (y - 4) 2 = 20 NextPreviousMain MenuMoving the Center Skip Tangents End

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intersect1 2 A line in the plane of a circle can intersect the circle in 1 or 2 points. exactly one A line that intersects the circle in exactly one tangent point is said to be tangent to the circle. The line and the circle are considered tangent to each other at this point of intersection. Write an equation for a circle with center (-4, -3) that is tangent to the x-axis. diagram A diagram will help. (-4, 0) (-4, -3) 3 perpendicular A radius is always perpendicular to the tangent line. (x + 4) 2 + (y + 3) 2 = 9 NextPreviousMain MenuMoving the CenterEnd

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The standard form equation for all conic sections is: Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 where A, B, C, D, E and F represent constants and where the equal sign could be replaced by an inequality sign. How do you put a standard form equation into graphing form? completing the square The transformation is accomplished through completing the square. NextPreviousMain MenuEndLast Viewed

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Graph the relation x 2 + y x + 4y + 13 = 0.13 F 1. Move the F term to the other side. x 2 + y x + 4y = -13 x-termsy-terms 3. Complete the square for the x-terms and y-terms. x x + y 2 + 4y = -13 x x y 2 + 4y + 4 = (x - 5) 2 + (y + 2) 2 = 16 x 2 + y x + 4y + 13 = 0 x 2 + y x + 4y = -13 x-termsy-terms 2. Group the x-terms and y-terms together x x + y 2 + 4y = -13 Graph the relation NextPreviousMain MenuCompleting the SquareEnd

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(x - 5) 2 + (y + 2) 2 = 16 (5, -2) center: (5, -2) 4 radius = 4 (x - 5) 2 + (y + 2) 2 = (5, -2) NextPreviousMain MenuCompleting the SquareEnd

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What if the relation is an inequality? x 2 + y x + 4y + 13 < 0 Do the same steps to transform it to graphing form. (x - 5) 2 + (y + 2) 2 < 4 2 This means the values are inside the circle. less thanradius The values are less than the radius. NextPreviousMain MenuCompleting the SquareEnd

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x 2 + y 2 + 6x - 2y - 54 = 0 Write x 2 + y 2 + 6x - 2y - 54 = 0 in graphing form. Then describe the transformation that can be applied to the graph of x 2 + y 2 = 64 to obtain the graph of the given equation. 1.x 2 + y 2 + 6x - 2y = 54 2.x 2 + 6x + y 2 - 2y = 54 3.( 6 / 2 ) = 3 ( -2 / 2 ) = -1 4.(3) 2 = 9 (-1) 2 = 1 5.x 2 + 6x y 2 - 2y + 1 = (x + 3) 2 + (y - 1) 2 = 64 7.(x + 3) 2 + (y - 1) 2 = center: (-3, 1) radius = 8 NextPreviousMain MenuCompleting the SquareEnd

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x 2 + y 2 + 6x - 2y - 54 = 0 x 2 + y 2 = 64 Write x 2 + y 2 + 6x - 2y - 54 = 0 in graphing form. Then describe the transformation that can be applied to the graph of x 2 + y 2 = 64 to obtain the graph of the given equation. x 2 + y 2 = 64 (x + 3) 2 + (y - 1) 2 = 64 x 2 + y 2 = 64 is translated 3 units left and one unit up to become (x + 3) 2 + (y - 1) 2 = 64. NextPreviousMain MenuCompleting the SquareEnd

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circle coefficientsx 2 y 2 equal The graph of a quadratic relation will be a circle if the coefficients of the x 2 term and y 2 term are equal (and the xy term is zero). PreviousMain MenuEnd

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